PAT甲级真题 1055 The World's Richest (25分) C++实现（数组排序，测试点超时问题）

题目

Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world’s wealthiest people. Now you are supposed to simulate this job, but concentrate only on the people in a certain range of ages. That is, given the net worths of N people, you must find the M richest people in a given range of their ages.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=10 ^ 5) – the total number of people, and K (<=10 ^ 3) – the number of queries. Then N lines follow, each contains the name (string of no more than 8 characters without space), age (integer in (0, 200]), and the net worth (integer in [-106, 106]) of a person. Finally there are K lines of queries, each contains three positive integers: M (<= 100) – the maximum number of outputs, and [Amin, Amax] which are the range of ages. All the numbers in a line are separated by a space.

Output Specification:

For each query, first print in a line Case #X: where X is the query number starting from 1. Then output the M richest people with their ages in the range [Amin, Amax]. Each person’s information occupies a line, in the format

Name Age Net_Worth
The outputs must be in non-increasing order of the net worths. In case there are equal worths, it must be in non-decreasing order of the ages. If both worths and ages are the same, then the output must be in non-decreasing alphabetical order of the names. It is guaranteed that there is no two persons share all the same of the three pieces of information. In case no one is found, output “None”.
Sample Input:

12 4
Zoe_Bill 35 2333
Bob_Volk 24 5888
Anny_Cin 95 999999
Williams 30 -22
Cindy 76 76000
Alice 18 88888
Joe_Mike 32 3222
Michael 5 300000
Rosemary 40 5888
Dobby 24 5888
Billy 24 5888
Nobody 5 0
4 15 45
4 30 35
4 5 95
1 45 50

Sample Output:

Case #1:
Alice 18 88888
Billy 24 5888
Bob_Volk 24 5888
Dobby 24 5888
Case #2:
Joe_Mike 32 3222
Zoe_Bill 35 2333
Williams 30 -22
Case #3:
Anny_Cin 95 999999
Michael 5 300000
Alice 18 88888
Cindy 76 76000
Case #4:
None

思路

直观的思路是，建立关于人的结构体数组，按财富值、年龄、姓名的优先级总体排序。每次查询都遍历该数组，输出前m个符合条件的即可。

测试点1用g++编译器有时会超时，换成clang++编译器总能通过（新技能get）：
在这里插入图片描述
这个方案最坏情况复杂度是O(kn)。

柳神的方法做了一些预处理。对整个数组排序后，若某个年龄出现次数超过100次，则后面的不做考虑，因为M最大为100，在财富值相等或递减的情况下，不可能输出同一个年龄100名以后的。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct Person{
    string name;
    int age;
    int worth;
};

bool cmp(Person &p1, Person &p2){
    if (p1.worth==p2.worth){
        if(p1.age==p2.age){
            return p1.name < p2.name;
        }
        return p1.age < p2.age;
    }
    return p1.worth > p2.worth;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<Person> p(n);
    for (int i=0; i<n; i++){
        cin >> p[i].name >> p[i].age >> p[i].worth;
    }
    sort(p.begin(), p.end(), cmp);
    for (int i=0; i<k; i++){
        int m, l, h;
        cin >> m >> l >> h;
        cout << "Case #" << i+1 << ":" << endl;
        int count = m;
        for (int j=0; j<n && count>0; j++) {
            if (p[j].age>=l && p[j].age<=h){
                cout << p[j].name << " " << p[j].age << " " << p[j].worth << endl;
                count--;
            }
        }
        if (count==m){
            cout << "None" << endl;
        }
    }
    return 0;
}